高一数学对数的运算公式 及讲解

回答的准确点

基本性质：
1.a^(log(a)(b))=b
2.log(a)(MN)=log(a)(M)+log(a)(N);
3.log(a)(M/N)=log(a)(M)-log(a)(N);
4.log(a)(M^n)=nlog(a)(M)
推导
1.这个就不用推了吧，直接由定义式可得(把定义式中的[n=log(a)(b)]带入a^n=b)
2.
MN=M*N
由基本性质1(换掉M和N)
a^[log(a)(MN)] = a^[log(a)(M)] * a^[log(a)(N)]
由指数的性质
a^[log(a)(MN)] = a^{[log(a)(M)] + [log(a)(N)]}
又因为指数函数是单调函数，所以
log(a)(MN) = log(a)(M) + log(a)(N)
3.与2类似处理
MN=M/N
由基本性质1(换掉M和N)
a^[log(a)(M/N)] = a^[log(a)(M)] / a^[log(a)(N)]
由指数的性质
a^[log(a)(M/N)] = a^{[log(a)(M)] - [log(a)(N)]}
又因为指数函数是单调函数，所以
log(a)(M/N) = log(a)(M) - log(a)(N)
4.与2类似处理
M^n=M^n
由基本性质1(换掉M)
a^[log(a)(M^n)] = {a^[log(a)(M)]}^n
由指数的性质
a^[log(a)(M^n)] = a^{[log(a)(M)]*n}
又因为指数函数是单调函数，所以
log(a)(M^n)=nlog(a)(M)
其他性质：
性质一：换底公式
log(a)(N)=log(b)(N) / log(b)(a)
推导如下
N = a^[log(a)(N)]
a = b^[log(b)(a)]
综合两式可得
N = {b^[log(b)(a)]}^[log(a)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
又因为N=b^[log(b)(N)]
所以
b^[log(b)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
所以
log(b)(N) = [log(a)(N)]*[log(b)(a)] {这步不明白或有疑问看上面的}
所以log(a)(N)=log(b)(N) / log(b)(a)
性质二：（不知道什么名字）
log(a^n)(b^m)=m/n*[log(a)(b)]
推导如下
由换底公式[lnx是log(e)(x),e称作自然对数的底]
log(a^n)(b^m)=ln(a^n) / ln(b^n)
由基本性质4可得
log(a^n)(b^m) = [n*ln(a)] / [m*ln(b)] = (m/n)*{[ln(a)] / [ln(b)]}
再由换底公式
log(a^n)(b^m)=m/n*[log(a)(b)]
--------------------------------------------（性质及推导 完 ）
公式三:
log(a)(b)=1/log(b)(a)
证明如下:
由换底公式 log(a)(b)=log(b)(b)/log(b)(a)   ----取以b为底的对数,log(b)(b)=1
    =1/log(b)(a)
还可变形得:
    log(a)(b)*log(b)(a)=1